How do you find the definite integral for: [(6x^2+2) / sqrt(x)] dx for the intervals [1, 5]?
2 Answers
May 23, 2017
Explanation:
"rewrite as"
(6x^2)/(x^(1/2))+2/x^(1/2)=6x^(3/2)+2x^(-1/2)
rArrint_1^5(6x^(3/2)+2x^(-1/2))dx
"integrate each term using the "color(blue)"power rule"
• int(ax^n)=a/(n+1)x^(n+1); n!=-1
=[12/5x^(5/2)+4x^(1/2)]_1^5
=(12/5. 5^(5/2)+4. 5^(1/2))-(12/5 .1+4.1)
=60sqrt5+4sqrt5-32/5
=64sqrt5-32/5~~136.71" to 2 dec. places"
May 23, 2017
Explanation:
Let,
We know that,
Enjoy Maths.!